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MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS
CRAIG A . T O V E Y
Georgia Institute
of Technology, Atlanta,
GA 30332
(Submitted February 1984)
I.
INTRODUCTION
How many times can the same number appear in Pascalfs triangle?
eliminating occurrences due to symmetry, (7 ) = (
(n\
occurrences of 1 = (
ln\
V
'
Xn
7
"
After
), and the uninteresting
;
J and n = f J, the answer to this question is not clear.
More precisely, if 1 < k < n/2, we say that ( -, 1 is a -proper binomial coefficient.
Are there integers that can be expressed in different ways as proper
binomial coefficients?
Enumeration by hand or with a computer program produces some cases, given
in Table 1. The smallest is 120, which equals
(?) - (?)•
Table 1.
Small Multiple Occurrences of Binomial Coefficients
INTEGER
BINOMIAL COEFFICIENTS
120
Cs")- (?)
210
c:). (V)
1540
(?)• (?)
3003
i\% ('/)• (?)
7140
(?). en
11628
24310
(?)• (T)
Cs7)- (T)
There is even an instance of a number, 3003, which can be expressed in three
different ways. No clear pattern emerges; the cases just seem to be sprinkled
among the binomial coefficients. We conjecture that, for any t, there exist
infinitely many integers that may be expressed in t different (proper) ways as
binomial coefficients.
356
[Nov.
MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS
Here we prove the conjecture for the case t = 2. The proof is constructive and depends in an unexpected way on the Fibonacci sequence.
I. THE CONSTRUCTION
We seek solutions to
an especially tractable situation because it leads to a second-order equation.
In particular, if (1) holds, then n(k + 1 ) = (n - k)(n - k - 1). Let
x = n - k - 1.
(2)
Then x(x + 1) = n{n - x) so n2 - xn - (x2 + x) = 0 and
x + V5x2 + kx
n=
,0>l
(3^
^
(since n is positive).
lutions to (1).
Integer solutions to (3) therefore lead to integer so-
Since 5x2 + kx is even if and only if x is even, this means we must find
integers x such that 5x2 + kx is a perfect square. Now x and 5x + 4 have no
common factors except possibly 2 or 4, so a natural slightly stronger condition
would be that both x and 5x + 4 be perfect squares. In other words, we need
to find integers z such that 5z2 + 4 is a perfect square. These are given by
the following lemma.
Lemma 1: Let Fj denote the Fibonacci sequence.
(F..x + F - + 1 )
2
Then, for all J,
- 5F| -4<-l)*.
Proof: A straightforward calculation (see, e.g., [2], pp. 148-149) shows
(.Fj + 1 + F..J*
- 5F/ = 4
^
+ F.Fj_1
- F/) - -4(F? + F^^
- F?+1),
which yields the result by induction.
The lemma tells us that for any j even, z = Fj gives the perfect square
+Fj+1)2.
5s 2 + 4 = {F._x
This completes the construction.
Theorem 1: Let Fj denote the Fibonacci sequence. Then, for any even j , there
exists a solution to (1), where x = FJ and n and k are given by (2) and (3).
Remark: Letting Lj denote the Lucas sequence as usual, we can write this solution as
F
jh
+F
I
,.
F L
J d -F!
,
Theorem 2: Theorem 1 gives all solutions to (1).
Proof: It follows from the preceding discussion that any solution to (1) corresponds via (2) to some integer x such that 5x2+ kx is a perfect square. Let
a (resp. b) be the number of times 2 divides 5x + 4 (resp. x).
If a > 2, then
1985]
357
MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS
b - 2 and, conversely, b > 2 implies a - 2. Since 5 ^ + 4 and x have no common
factors except (possibly) 2 or 4, (5x + 4)/2 a is a perfect square, as is x/2b.
Therefore, a + & is even, so a and 2? are both even or both odd. In the former
case, x and 5x + 4 are perfect squares. We claim this leads precisely to the
class of solutions given by Theorem 1. In the latter case, it follows that
a = b = 1. Thus, we seek integers s such that 5s 2 + 2 is a perfect square. We
further claim that no such integers exist. The two claims can be shown to follows from the general theory of the so-called Pell equation (see, for example,
[1] for the first claim, and [3, pp. 350-358] for the second claim). For completeness, we give a simple proof that does not rely on the general theory.
Let {An} denote any sequence of positive numbers satisfying the recurrence
Tne
An + An+1 = An+2'
argument from Lemma 1 shows that, for all n,
+
(An-i
A
n+0
~ ~>An - 4(An_1 + An_1An
- An)
+ An
= -\An
+ 2)
+ 5An + 1 .
2
2
Therefore, given any solution s, y to 5s + k - y , we can construct smaller
solutions by setting
y
A
A
i
S
~ >
A
~
i-1
z
"
2
i +1 ~
and extending the sequence {An}
A
n
+
A
n+i
=
-
A
A
5
y
+
2
z
'
backward according to the recurrence
A
n+2*
[The solutions will be z = Aj, y = AJ-_1 + Aj+li
where j = i (mod 2), with
\k\ = h\An + AnA2+1
- A2+1\,
for all n.]
Now, let (s, y) be any integer solution to 5z2 + 4 = z/2.
^
= 2
Then extend {An}
If z > 3, then
and
^- + 1
=
— o —
Set
(an integer).
backward to get the solution corresponding to A^_
., ^ z/5 - z ^ .
.61s ^
«
^ ^i-i
=
and A._
.
y - % ^ zy/5 + .5 - z ^ _0
— 9 —
2
-72s,
whence .28s < Ai_2 < .39s. Therefore, if Ai ^ 3, the solution corresponding to
i4^_2 is smaller. Repeatedly extend {An} backward until Aj < 3. Since the only
such integer solution is (1, 3 ) , s must have been a Fibonacci Number. This
verifies the first claim. The second claim, that 5s 2 + 2 = y2 has no solutions,
follows immediately from the fact that y2 = 2 (mod 5) has none.
REFERENCES
1.
2.
3.
358
I.Gessel. Advanced Problem H-187. The Fibonacci
Quarterly
10 (1972):417419.
G. H. Hardy & E. M. Wright. An Introduction
to the Theory of Numbers.
Oxford: Oxford University Press, 1959.
J. Uspensky & M. Heaslet. Elementary
Number Theory.
New York: McGraw-Hill,
1939.
[Nov.